How is finiteness of solutions (zero-dimensionality) related to Krull's dimension?

Question

I have encountered a lot the concept of zero-dimensional ideal:

Let $k$ be a field. An ideal $I\subseteq k[x_1,...,x_m]$ is said to be zero-dimensional if its zero set $Z(I)$ has a finite number of elements.

In these notes, it is shown that an ideal is zero-dimensional if and only the dimension of $k[x_1,...,x_m] / I$ as a $k-$vector space is finite, and this is usually given as a definition. However, I'm concerned with the relation of this concept with Krull's dimension.

Krull's dimension is defined as the supremum of the lengths of all chains of prime ideals. According to this post in MSE, $I$ is zero-dimensional if and only if its quotient ring $k[x_1,...,x_m] / I$ has Krull's dimension zero. My first question is:

Is this the reason why the word dimension is involved in the zero-dimensional definition?

This is not a primarily opinion-based question. I'm just asking if the word dimension is involved in the definition is related to Krull's dimension in some way or not.

My second question is related with the second link I shared (the MSE post). It says that $k[x_1,...,x_m] / I$ has Krull's dimension zero if and only if $I$ is contained in finitely many prime ideals. However, what if $I$ is contained in at least two different prime ideals that have a containment relation? wouldn't the Krull's dimension of $k[x_1,...,x_m] / I$ be non-zero in this case?

Thanks a lot for your time, I really hope I can receive some help with these concepts.

Answer 1

For your first question: the reason it is called a zero dimensional ideal is very intuitive: the zero set of such an ideal consists of finitely many isolated points in affine space. One thinks of sparse points as being zero dimensional things, as opposed to curves, surfaces, hypersurfaces etc.

You can relate Krull dimension to the notion of zero dimensional ideal; I am not sure which use of the word dimension appears in the literature first, but both uses refer to the same geometric idea. If $Z$ is a closed set in affine $n$ space, then then one has a notion of topological dimension of $Z$: the maximal number of inclusions of a chain of irreducible closed subsets of $Z$. If $I$ is the ideal corresponding to $Z$, then the topological dimension of $Z$ turns out to be the Krull dimension of the ring $k[X_1, ... , X_n]/I$. So as you mentioned, saying $I$ is zero dimensional is the same as saying the ring $k[X_1, ... , X_n]/I$ is zero dimensional, i.e. has no inclusions among prime ideals.

For your second question, it turns out that can't happen. Let me do the case where $k$ is algebraically closed. If $I \subseteq \mathfrak p_1 \subsetneq \mathfrak p_2$, then $I$ will be contained in infinitely many maximal ideals. Indeed, $Z(\mathfrak p_2)$ is strictly contained in $Z(\mathfrak p_1)$, so $Z(\mathfrak p_1)$ must be an infinite set (the $Z(\mathfrak p_i)$ are irreducible; in a topological space where one point sets are closed, each singleton set is irreducible, so any nonsingleton irreducible set must be infinite). Hence $Z(I)$ is an infinite set, i.e. $I$ is not zero dimensional. And, for each point $P$ in $Z(I)$, you will have $I$ contained in the ideal corresponding to $P$, and ideals corresponding to points are maximal.

For the case where $k$ is not algebraically closed, you actually need $I$ to satisfy the following property (usually given as the definition of zero dimensional ideal, but I am not sure if it the one you are using): an ideal $I$ of $R = k[X_1, ... , X_n]$ is zero dimensional if its zero set in $K^n$ for every field $K$ containing $k$ is finite. In this case, let $K$ be an algebraic closure of $k$, then $S = K[X_1, ... , X_n]$ is integral over $R$. The inclusion map $R \rightarrow S$ induces a ring homomorphism $\phi: R/I \rightarrow S/IS$. The zero set of $I$ in $K^n$ is the same as the zero set of $IS$, so by the algebraically closed case, $\textrm{Dim } S/IS = 0$. By the going up theorem, if $A$ is a subring of $B$ with $B$ integral over $A$, then $\textrm{Dim } A \leq \textrm{Dim } B$. Since $S$ is integral over $R$, $S/IS$ is integral over the image of $\phi$, so $$\textrm{Dim } R/I \leq \textrm{Dim } \phi(R/I) \leq \textrm{Dim } S/IS = 0$$ Actually the first $\leq$ is an equality, because $\phi$ is injective (that is to say, $I = IS \cap R$), but that requires some commutative algebra machinery to prove quickly, and you don't need it here.

Answer 2

This is just adding to D_S's answer, saying the last paragraph algebraically just for clarity.

If $R$ is a domain, f.g. over a field $k$, which is not itself a field, then $R$ has infinitely many maximal ideals.

I don't think you can avoid at least some tricky algebra here. (Which D_S slips into topological language - this is a really good habit, I think! But when you are learning the subject, like I am, it seems important to keep track of what is going on behind the scenes.)

Pf: By Noether normalization, there is a subring of $R$ which is isomorphic to $k[t]$, and $R$ is a finite module over this $k[t]$. Now the going up (or down?) theorem implies that for each prime $P$ of $k[t]$, there is a prime $Q$ of $R$ which restricts to $P$ ($Q \cap k[t] = P$). We can write down infinitely many maximal ideals of $k[t]$.